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Question-187559




Question Number 187559 by Mingma last updated on 18/Feb/23
Answered by cortano12 last updated on 19/Feb/23
 ((sin x)/(sin y)) .((cos y)/(cos x)) = (1/3)⇒((tan x)/(tan y))=(1/3)   3tan x=tan y  ⇒tan (x+y)=((4tan x)/(1−3tan^2 x)) =± (√(15))  ⇒2sin x=sin y ; 2cos x=3cos y  ⇒4=sin^2 y+9cos^2 y  ⇒4=sin^2 y+9−9sin^2 y  ⇒sin^2 y=(5/8) ∧ cos^2 y=(3/8)  ⇒tan y=± (√(5/3))
$$\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{y}}\:.\frac{\mathrm{cos}\:{y}}{\mathrm{cos}\:{x}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\frac{\mathrm{tan}\:{x}}{\mathrm{tan}\:{y}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\mathrm{3tan}\:{x}=\mathrm{tan}\:{y} \\ $$$$\Rightarrow\mathrm{tan}\:\left({x}+{y}\right)=\frac{\mathrm{4tan}\:{x}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}}\:=\pm\:\sqrt{\mathrm{15}} \\ $$$$\Rightarrow\mathrm{2sin}\:{x}=\mathrm{sin}\:{y}\:;\:\mathrm{2cos}\:{x}=\mathrm{3cos}\:{y} \\ $$$$\Rightarrow\mathrm{4}=\mathrm{sin}\:^{\mathrm{2}} {y}+\mathrm{9cos}\:^{\mathrm{2}} {y} \\ $$$$\Rightarrow\mathrm{4}=\mathrm{sin}\:^{\mathrm{2}} {y}+\mathrm{9}−\mathrm{9sin}\:^{\mathrm{2}} {y} \\ $$$$\Rightarrow\mathrm{sin}\:^{\mathrm{2}} {y}=\frac{\mathrm{5}}{\mathrm{8}}\:\wedge\:\mathrm{cos}\:^{\mathrm{2}} {y}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{tan}\:{y}=\pm\:\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\: \\ $$
Commented by Mingma last updated on 19/Feb/23
Great man!

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